18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)

19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được 

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)

Ta có ;

 \(\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=\frac{\left(a+b\right)\left(a-b\right)+\left(b+c\right)\left(b-c\right)+\left(c+a\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

17. Xét vế trái ; 

\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{a}{\left(b-1\right)\left(b^2+b+1\right)}+\frac{b}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{a}{-a\left(b^2+b+1\right)}+\frac{b}{-b\left(a^2+a+1\right)}=\frac{-1}{b^2+b+1}+\frac{-1}{a^2+a+1}\)

\(=\frac{-\left(a^2+a+1+b^2+b+1\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}=\frac{-\left[\left(a+b\right)^2-2ab+3\right]}{a^2b^2+ab\left(a+b\right)+a^2+b^2+ab+2}\)\(=\frac{2\left(ab-2\right)}{a^2b^2+\left(a^2+2ab+b^2\right)+2}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)

10. a) 

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow\left(a+b\right)\left(x^4+y^4\right)=ab\left(x^2+y^2\right)^2\Leftrightarrow\left(bx^2-ay^2\right)^2=0\Leftrightarrow bx^2=ay^2\)

b) Từ \(ay^2=bx^2\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\); \(\frac{y^{2008}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

25. Ta có \(\left(ax+by+cz\right)^2=0\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(abxy+bcyz+acxz\right)\)

Xét mẫu số của P : \(bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)

\(=y^2bc-2bcyz+bcz^2+acx^2-2xzac+acz^2+abx^2-2abxy+aby^2\)

\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(abxy+xzac+bcyz\right)\)

\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)

\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)

\(\Rightarrow P=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{2007}\)

8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)

8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)

câu 1 là :từ a/x + b/y + c/z =0 suy ra (ayz+bxz+cxy)/xyz =0 suy ra ayz+bxz+cxy=0 (1)

vì x/a + y/b + z/c =1 (gt) suy ra (x/a + y/b + z/c )^2 = 1^2 . suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(xy/ab + yz/bc + xz/ac) =1

suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2[(ayz+bxz+cxy)/abc = 1 (2)

Từ (1) và (2) suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =1 (đpcm)

câu 3 98

Tks pạn nhìu @!

1) Thay xyz = 1  , ta có : 

 \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)

\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)

2) Phân tích A thành nhân tử được \(A=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

Vì a + b + c = 0 nên A = 0

3) Phân tích  A thành  \(\frac{\left(b-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)